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Alex787 [66]
3 years ago
15

Unlike most real bulbs, the resistances of the bulbs in the questions below do not change as the current through them changes. A

ll bulbs considered in this problem are identical. Assume all batteries are ideal (they have no internal resistance) and connecting wires have no resistance.
1. The current passing through the brighter

bulb is larger.

2. Not enough information is given.

3. The current passing through the brighter

bulb is smaller.

4. The current passing through both bulbs

are the same.
Physics
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

Option 1 is correct.

The current passing through the brighter bulb is larger.

Explanation:

The brightness of the bulb is determined by the power, I²R

And since they all have equal resistances, the only factor different that could result in more or less power is the current, I through the bulb.

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Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of
mart [117]

Answer:

The value is  B =  3.33 *10^{-6} \  T

Explanation:

From the question we are told that

  The distance of separation is  d = 0.6 \  m

  The current on the one wire is I_1 =  9 \  A

  The current on the second wire is I_2 =  4 \ A

Generally the magnitude of the field exerted between the current carrying wire is

        B  =  B_1 - B_2

Here B_1 is the magnetic field due to the first wire which is mathematically represented as

         B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}

Here d_1 is the distance to the half way point of the separation and the value is  

    d_1 =  0.3 \  m

B_2 is the magnetic field due to the first wire which is mathematically represented as

         B_2  = \frac{\mu_o * I_2 }{2 \pi * d_2}

Here d_2 is the distance to the half way point of the separation and the value is  

    d_2 =  0.3 \  m  

This means that d_1 = d_2 = a =  0.3

So

     B =  \frac{\mu_o * I_1 }{2 \pi * d_1}  -  \frac{\mu_o * I_2 }{2 \pi * d_2}

=>  B =  \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }

=>  B =  \frac{  4\pi * 10^{-7}  * (9- 4)}{2 * 3.142  *0.3 }

=>  B =  3.33 *10^{-6} \  T

5 0
3 years ago
A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc
guajiro [1.7K]

Answer:

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

a_o=a\times e^{-kt}

where,

k = rate constant  

t = age of sample

a_o = let initial amount of the reactant  

a = amount left after decay process  

We have :

a_o=x

a=58\%\times x=0.58x

t = 95 s

0.58x=x\times e^{-k\times 95 s}

\k= 0.005734 s^{-1}

Half life is given by for first order kinetics::

t_{1/2}=\frac{0.693}{k}

=\frac{0.693}{0.005734 s^{-1}}=120.86 s

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

3 0
3 years ago
This what i actually look like
Natali5045456 [20]

you're so beautiful!

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4 0
3 years ago
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Ya because they’re are both 50 liters
7 0
2 years ago
What is the magnitude of the average velocity of the hummingbird between 2 sec and t 4 sec <br>​
Arte-miy333 [17]
Between magnitude of the average 4sec
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